Is there a collectionwise normal topological vector space which is not paracompact?

Let P2(r) denote the vector space?

• (a) Let P2(R) denote the vector space of real polynomial functions of degree less than or equal to two and let B := [p0, p1, p2] denote the natural ordered basis for P2(R) (so pi(x) = xi). Define f 2 P2(R) by f(x) = 5x2 − 2x + 3. Write f as a linear combination of the elements of B. Compute the coordinate vector fB of f with respect to B. Define h1, h2, h3 e P2(R) by h1(x) = 7x2 + 3x + 4, h2(x) = x2 + 2x + 1 and h3(x) = x2 − 1. Define C := [h1, h2, h3]. Assuming C is an ordered basis for P2(R), construct the change of coordinate matrix, A, which converts C-coordinates to B-coordinates. Compute A^−1. Let M denote the change of coordinate matrix that converts B-coordinates to C-coordinates. How are A and M related? Explain why the calculations you have performed show that C is a basis for P2(R). Compute the coordinate vector fC of f with respect to C.

• Answer:

  this isn't really a hard question, the difficulty lies in "untangling" what is meant by it all. first of all, B is a really friendly basis, it's just {1,x,x^2}. thus the "B-coordinates" of a vector in P2(R), are just the coefficients of the polynomial. so if f(x) = 5x^2 - 2x + 3, then [f]B = [3,-2,5]B. (this is just a fancy way of saying: 5x^2 - 2x + 3 = (3)(1) + (-2)x + (5)(x^2), easy, huh?) C isn't quite so friendly. in fact, it's not even clear it IS a basis, just yet. for example, we don't know right off the bat what dim(span(C)) is (except that it's at most 3). but we can still express the elements of C in terms of B: [h1]B = [4,3,7]B, since h1(x) = 4(1) + 3(x) + 7(x^2). [h2]B = [1,2,1]B (why?) [h3]B = [-1,0,1]B now, if we "pretend" C is in fact a basis, then the "C-coordinates" of h1 are [1,0,0]C, since h1 = (1)h1 + (0)h2 + (0)h3. similarly, [h2]C = [0,1,0]C, and [h3] = [0,0,1]C. now if A is a matrix that turns C-cooridnates into B-coordinates, then: [* * *][1]....[4] [* * *][0]....[3] [* * *][0] = [7] this tells us the first column of A is (4,3,7). similarly, the second column of A must be [h2]B, and the third column of A is [h3]B, so A = [4 1 -1] [3 2 0 ] [7 1 1 ] is A invertible? well, a "sneaky way" to find A^-1 is to express the B-vector basis in terms of the C-vectors. that is, find aj,bj,cj (j = 1,2,3) so that: 1 = a1h1 + b1h2 + c1h3 x = a2h1 + b2h2 + c2h3 x^2 = a3h1 + b3h2 + c3h3 a little playing around with h1,h2 and h3 gives: 1 = h1/8 - 3h2/16 - 11h3/16, so the first column of "A^-1" (we don't actually know it's a true inverse, yet) will be (1/8, -3/16, -11/16). more playing around gives: x = -h1/8 + 11h2/16 + 3h3/16. so our second column is (-1/8,11/16,3/16). continuing we find that: x^2 = h1/8 - 3h2/16 + 5h3/16, so the third column is (1/8,-3/16,5/16). we can now verify we have a true inverse by multiplying: [4 1 -1][1/8 ...-1/8..... 1/8] [3 2 0 ][-3/16 11/16 -3/16] [7 1 1 ][-11/16 3/16 .5/16] = [1 0 0] [0 1 0] [0 0 1]. in the course of doing this, we showed that M = A^-1. but this means that "A" is a bijective linear transformation on P2(R), which means that: that the columns of A are linearly independent. since we have 3 linearly independent vectors in the set C = {h1,h2,h3}, and since dim(P2(R)) = 3, we have that C spans P2(R) as well, and is thus a basis for P2(R). of course, to find [f]C, we need only compute M([f]B) = (A^-1)([f]B] = [1/8 ...-1/8..... 1/8][ 3]....[...5/4] [-3/16 11/16 -3/16][-2]....[-23/8] [-11/16 3/16 .5/16][ 5] = [..-7/8] as a final (paranoia) check, we compute that: (5/4)h1 - (23/8)h2 - (7/8)h3 = (35/4)x^2 + (15/4)x + 5 - (23/8)x^2 - (23/4)x - (23/8) - (7/8)x^2 + 7/8 = (70/8 - 23/8 - 7/8)x^2 + (15/4 - 23/4)x + (40/8 - 23/8 + 7/8) = 5x^2 - 2x + 3 = f(x)